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Question

Two non - integer roots of the equation $$(x^{2}+3x)^{2}-(x^{2}+3x)-6= 0$$  
are 


A
12(3+11),12(311)
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B
12(3+7),12(37)
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C
12(3+21),12(321)
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D
none of these
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Solution

The correct option is C $$\frac{1}{2}(-3+\sqrt{21}),\frac{1}{2}(-3-\sqrt{21})$$
Equation is $$(x^{2}+3x)^{2}-(x^{2}+3x)-6=0$$
Put $$x^{2}+3x=y$$
Therefore,
$$y^{2}-y-6=0$$
$$(y+2)(y-3)=0$$
$$y=-2,3$$
When $$y=-2$$
$$x^{2}+3x=-2$$
$$x^{2}+3x+2=0$$
$$(x+1)(x+2)=0$$
$$x=-1,-2$$
When, $$y=3$$
$$x^{2}+3x-3=0$$
$$x=\dfrac{-3\pm \sqrt{9+12}}{2}$$
$$x=\dfrac{1}{2}(-3+\sqrt{21}), \dfrac{1}{2}(-3-\sqrt{21})$$

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